Physics Concept Questions: Kinetic energy and momentum problems

Write out the force of each particle in the particle group: fi = FJI + F outside I, where FJI is the force of the j-th particle on it, and F outside I is the external force.

By definition: FJI=-FIJ.

Now look at the change in the total energy of the system, δ = δ i, where δ i is the energy change of the ith particle. δ i=fi·δsi=fji·δsi+ fji·δsi

= F-Outer I·ΔSi+ (FJI·ΔSi)= F-Outer I·ΔSi+ FJI· (δsi-δsj), so if there is a relative shift between the ith and jth particles, i.e., (δsi-δsj)≠0

From the relation of δ, we can get that fji is doing work.

Looking at the change in the total momentum of the system, δ = δ i= fiδti= (f outside i + fji)δti

After opening, we still get a formula similar to the change of kinetic energy: δ = F outside IδTi+ FJI(δti-δtj), but the time experienced by all particles in this process is the same δTi=δtj=δt, so the impulse of the internal force is 0.

To put it simply, when calculating total momentum, f*t is used to calculate the total momentum, and the time experienced by different objects is the same, and when calculating total kinetic energy, f*s is used, and the displacement experienced by different objects in this time is different.

That is, the formula you listed ek=f1s+f2s+f1s+f2s, should be changed to: ek=f1s1+f2s2+f1s1+f2s2

And S1 and S2 can be different.

Quite simply, suppose two metal balls A and B of mass m are connected by a compressed spring (of negligible mass) connected by a neutral mass and are stationary on a smooth horizontal plane.

On the whole, the total kinetic energy of this system (l) is 0 at this time. If the spring is suddenly loosened at this point (you can imagine that a string is broken between the two balls, and the mass is negligible), the two balls must have an opposite velocity v, the kinetic energy of the whole system is mv*v, and the momentum is still 0

Think of the above process as a theoretical experiment, and the results prove your argument.

When there is internal energy, the transformation of energy is realized, so change the kinetic energy! Brakes are an example of this! And momentum can be used macro**, and the internal energy belongs to the internal forces inside the system, so it does not affect the momentum!

The work done by selecting B and D bullets to overcome resistance is partly to overcome friction, so it is greater than the kinetic energy obtained by the wooden block, so A is wrong, and C is wrong.

First of all, the horizontal resultant external force is 0Momentum is conserved, mv1=mv2, and secondly, energy is conserved, and the gravitational potential energy of the ball is converted into kinetic energy of both.

mgl(1-cosa) = mv2 2 2 + mv1 2 2 easy to solve v2 = root number (2mgl(1-cosa) divided by (m+m)).

1. Because it is in a smooth plane and there is no external force in the horizontal direction, the momentum in the horizontal direction of the ball + car system is conserved. Therefore, the horizontal momentum of the whole system should be in a state of zero sum at all times.

2. Since there is no process of conversion of mechanical energy into other energy in the whole process, mechanical energy is conserved. Therefore, at all times, the total mechanical energy of the whole system should be equal to the sum of the gravitational potential energy in the initial state.

At the lowest point, the speed of the car is v, and the ball speed is v to obtain the following equation:

mv = mvmv 2 2 = mg(l-cos *l)-mv 2 2 solution gives v = root number (2mgl(1-cosa) divided by (m+m)).

If the resultant force of the system is 0, the centroid acceleration is 0, and if the centroid velocity is 0, the centroid position remains unchanged.

Let the distance of the boat retreat x, then -mx+n*m*(l-x)=0 gives x=nml (m+nm).

How do you calculate the resistance of water to a boat?

b For this kind of multiple-choice question, a special method can be used, and the quality of the person and the car is the same as mAt the beginning, the momentum of man and c is always 0, and then man jumps to the left with v, naturally due to the conservation of momentum, c moves to the right with v; Then the person goes up to B, and the total momentum of the two is MV, to the left; The person jumps to the left with the ground v, because the momentum is conserved, b can only be stationary, at this time the person jumps on a, so the momentum is conserved, and the common velocity of him and a can only be v 2, so b is chosen

Let the velocity of the shell leave the muzzle (to the ground) is v0, and the recoil velocity of the gun body is v1 (horizontal).

then v 2 (v0*sin ) 2 (v0*cos v1) 2 ...Equation 1

Taking the shell and the gun body as a system, the partial momentum of the system in the horizontal direction is conserved, get.

m m) * v1 m v0 * cos, m is the mass of the cannon (barrel and shells).

i.e. v0 (m m)*v1 (m *cos).Equation 2

The above two forms are combined, and they are obtained.

v^2＝[（m－m）*v1 / （m *cosα）]2*（sinα）^2＋^2

v^2＝[（m－m）*v1 / （m *cosα）]2*（sinα）^2＋^2

v^2＝v1^2 * m－m)^2＋m*(2m－m)*(cosα)^2 ] / m^2

The recoil velocity of the cannon is v1 v * m root number [ (m m) 2 m * (2m m) * (cos) 2 ].

m v / (m－m*sinα)

Taking the cannon and the shell as a system as the research object, the net force of the system in the horizontal direction is zero, the momentum in the horizontal direction is conserved, and the velocity direction of the gun body is in the positive direction, and it is set to vx, then there is mvx-mvcos = 0, then vx = mvcos m

The velocity of the shell to the ground is the VCOS -V cannon.

Conservation of momentum in the horizontal direction:

MV cannon = m (VCOS -V cannon).

Seek: V cannon = MVCOS (m+m).

mvcosα/(m+m)

I've already considered the relative rate, if the absolute rate is mvcos m. Isn't that right?

a. After diving, taking the boat as the reference system, the initial velocity of the boat is taken as 0, and the backward velocity of the boat is set v1mv=(m+m)v1

After b, after diving, take the boat at this time as the reference system, and the speed is still regarded as 0, and the boat speed v2mv=mv2

According to the speed relation of the reference frame taken twice.

Final boat velocity v=vo+v2-v1=vo+mv[1 m-1(m+m)]momentum part of the recoil ** model, as long as we talk about the ejection velocity, it must refer to the velocity relative to the individual separated from it, and it is impossible to say that it is other frame of reference, as ls thinks, both v are relative to the ground velocity, then this part is not a problem.

The velocity should not change, the momentum of the two children is in opposite directions, canceling each other, note that in the conservation of momentum, the velocity refers to the absolute velocity, and the absolute velocity of the two children is the same.

1l error: Because the force object is different when changing twice, the first time is m+m, and the second time is m

Therefore, the final velocity is.

V end = V0-MV (m+m)+mv M

Complement Reroger.

V end = V0-MV (m+m)+mv M

It's easy to misunderstand when writing in a row.

According to the conservation of momentum mv0=(m+m)v, v can be calculated;

The condition for m not to fall is that it is stationary relative to m before reaching a, if, when m reaches point a, it is at the same speed as m, according to the kinetic energy theorem: ml=1 2*mv 2, bring in the value of v, you can find: v0=(2 l)*(m+m) m under the root number.

μml=(1/2)*mv0^

v0 = 2 ml m under the root number

So v0<=2 ml m under the root number is in line with the topic.